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emmab  
#1 Posted : Wednesday, 2 September 2015 10:47:15 PM(UTC)
emmab

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Background
Vinegar contains a small percentage of ethanoic acid, CH3COOH. This experiment intends to find out the concentration of the vinegar against a standard solution of sodium hydroxide solution of concentration of 0.995 M through acid-base titration. The equation of the reaction between ethanoic acid and sodium hydroxide is as follows:

CH3COOH + NaOH = CH3COONa + H2O

The end point of the titration process can be determined and the amount of reactants used can be measured. Using the values obtained from the titration, and also the chemical equation as reference, the concentration of ethanoic acid in the vinegar can be determined through stoichiometric calculations.

Phenolphthalein indicator solution will be used in this acid-base titration.

A 25 mL volume of 0.995M sodium hydroxide solution was pipetted into a flask and several drops of phenolphthalein indicator was added. Using a burette, vinegar was slowly added until the indicator turned permanently clear. The average titre (volume of vinegar added) was 28.99mL.
Determine the concentration of ethanoic acid in the vinegar.

Data processing/analysis
From the equation of the chemical reaction between diluted vinegar solution (ethanoic acid, CH3COOH solution) and sodium hydroxide, NaOH solution,

CH3COOH + NaOH = CH3COONa + H2O

We can deduce that 1 mol of ethanoic acid is needed to neutralise 1 mol of sodium hydroxide. In other words, the amount of moles ethanoic acid used in the reaction is equal to amount of moles of sodium hydroxide used.

Therefore, to find the concentration of the diluted ethanoic acid solution,

Amount of moles of CH3COOH = amount of moles of NaOH

Macid x Vacid = Mbase x Vbase
Macid x 28.99 = 0.995 x 25
Macid = 25x0.995/28.99=0.858M

To find concentration of vinegar prior to dilution, Mvinegar,

Amount of moles CH3COOH before dilution = amount of models of CH3COOH after dilution
Mvinegar x V1 = Macid x V2

If you have any questions, please post here....
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